Hvac Barometric Pressure Relief Damper

Barometric Pressure Relief Damper William A. Greco [email protected] Trident3 June 2009

Page 1 of 9

Summary This report will show that simple* barometric pressure relief** dampers can be statically calibrated accurately by the use of mathematical principles. As no means of calibration markings are included on balance weights on Barometric Pressure Relief Dampers, the proper calibration to initially set a given damper is usually confined to turning on the system, manually and dynamicaly setting the damper by the balancer while the system is in operation. * Simple in that only gravitational and pressure control is available and no electronic means of operation is provided for damper operation. ** Relief = indicating that the damper must open at a preset minimum value.

Main Most true Barometric Pressure Relief Dampers have an axle that is off center with respect to thier perifery. Some Barometric Pressure Relief Dampers are constructed with an axle that is centered and require a balance weight that can be radially offset. However dampers of this type should be avoided as their actual operation is in doubt. A Barometric Pressure Relief Damper with an equal amount of pressure on both sides of the axle cannot produce the forces of velocity pressure to cause an imbalance that percipitates dynamic movement.

Figure-1 shows a Barometric Pressure Relief Damper that has equal blade areas on both the top and bottom which corresponds to an equal velocity pressure on each side of the damper axle.

Barometric Pressure Relief Damper William A. Greco [email protected] June 2009

Page 2 of 9 Trident3

Figure 2 indicates a Barometric Pressure Relief Damper operated by pendulum action of the attached weight, which is set to a desired angle by means of an adjustable collar.

The shaded part of the circle in figure-3 represents the smaller area of a Barometric Pressure Relief Damper with an offset axle.

Barometric Pressure Relief Damper William A. Greco [email protected] Trident3 June 2009

Page 3 of 9

Refering to figure-3 shown on page-2 , the areas A1 and A2 of the circle can be calculated thus:

C = 2 2hr-h 2

equation-1

And

 4h 2   C 2  A2 =   0.392   2   3   4h  good up to a semi-circle

equation-2

Substituting equation-1 into equation-2:

 2  4h   2 2hr-h A2 =    4h 2  3   



2

2

  0.392 

equation-3

  

And it follows that:

 2 2hr-h 2    4h A1 =  r 2     4h 2  3    2



2

  0.392    

eqution-4

Barometric Pressure Relief Damper William A. Greco [email protected] June 2009

Page 4 of 9 Trident3

The centroid of a plane surface is a point that corresponds to the center of gravity of a very thin homogenous plate, at that plates center of area. The force per unit area acts through the centroid. The damper torque on each side of the axle is product of the distance from the axle center line to the centroid times the force per unit area on the plate times the plate area normal to the direction of flow. After the areas have been ascertained it becomes necessary to calculate the centroid of each area (A1 and A2). Refering to figure-4, the centroid will act at some point along h1 and h2 at the center of the semi-circular areas.

To calculate centroid values h1 and h2 the following equations can be used:

 C 3  h1 = equation-5a 12A1

and

 C 3  h2 = 12A 2

equation-5b

Although equations 5a and 5b are recommended for their pure simplicity the following equations can also be used. 2 3  2   r sin 1  h1 =      3   A1 

equation-6a

2 3  2   r sin  2  h2 =     equation-6b  3   A2 

where: A1 or 2 = area

r = damper radius

 4r sin 3  1  h1 =    61 -3 sin 2 1 

equation-7a

 4r sin 3  2  h2 =   equation-7b  6 2 -3 sin 2 2 

Barometric Pressure Relief Damper William A. Greco [email protected] June 2009

Page 5 of 9 Trident3

Available Force to Provide Momentum The theory of air resistance deduced from the principles of mechanics, are given in the Philosaphiae Naturalis Principia Mathematica of Sir Isaac Newton . He stated that the forces acting between a solid and a fluid are the same whether the body moves with a velocity through the fluid or the fluid moves against a body at rest. This is valid for bodies of similar shape. The following statments say that the forces acting on two geometrically similar bodies which move in fluids with different densities are proportional to- (these statements will not be required until after the damper’s first instant of movement, in the example given here we assume that the air pressure forces are known). a) the square of the velocity, b) b) the square of the linear dimension of the body, and c) the density of the fluid. To solve for the torque at the first instant of movement we can use the figure –5 on this page and equation-8 on page-6 to determine the force avilable to open the barometric damper:

Barometric Pressure Relief Damper William A. Greco [email protected] June 2009

T=

A1h1 

 P 

27.72

-

Page 6 of 9 Trident3

A2 h 2

-  P 

27.72

= Torque in inch pounds

equation-8

where: P = Total Pressure at the point in the duct (inwg) Total Pressure = (Static Pressure + Velocity Pressure) - friction up to the damper location 1 psi = 27.72 inches of water gage (inwg)   distance from center line of damper to axle position A1 = Area of larger part of damper h 1  distance of centerline to A1 centroid A2 = Area of smaller part of damper h 2  distance of centerline to A2 centroid Example-1: Assume a 16 inch diameter barometric damper with an axle offset (Y) of 2 inches. (see figure-6), assume a system with a total pressure of 2.38 inwg at the damper.

Barometric Pressure Relief Damper William A. Greco [email protected] June 2009

Page 7 of 9 Trident3

To find C (the chord) of the semi-circle we first use equation-1. C = 2 2hr-h 2 equation-1 where: C = the length of the chord also the length of the axle within the damper h = the height from the chord to the perimeter of the damper r = the radius of the damper 2

C = 2 2 6 8   6   15.492 inches Next the areas of A1 and A2 must be calculated. First calculate A2 by the use of equation-3 found on page-3 of this report: 2   2 2 2hr-h   4h   A2 =   0.392 equation-3    4h 2  3      where: A2 = the area of the dampers smaller semi-circle h = the height from the chord to the perimeter of the damper r = the radius of the damper



2

46  2

A2 =

3



2   2  2 2 6 8  8      0.392 = 68.871 in 2 2 4 6 

To find A1 which is the area of the larger part of the damper subtract A2 from the total area of the damper:

A1 =  r 2 - A2 3.14 82  - 68.871 = 132.2 in 2 Next we must find the centroids of A1 and A2 parts of the damper. See page 7.

Barometric Pressure Relief Damper William A. Greco [email protected] June 2009

Page 8 of 9 Trident3

C3   C3  h1 = equation-5a and h2 = equation-5b 12A1 12A2 where: C = chord length (axle length) = 15.492 inches h1 ,h 2 = the centroid distances measured from the damper's centerline A1,A2 = The area's of the respective damper segments 15.4923  h1 = = 2.344 inches from the centerline 12 132.2  15.4923  h2 = = 4.499 inches from the centerline 12 68.87 

Employing equation-8 from page-6 the available torque at the damper is found. A1h1   P  A2 h 2 -  P  T= = Torque in inch pounds equation-8 27.72 27.72 132.2 2.34  2 2.38 68.94.5 - 22.38 T= = 0.318 inch pounds available for movement 27.72 27.72 A 1” long rod attached to the axle (see figure-2) with a weight weighing 0.319 pounds will keep the damper from opening. A pressure above the setting of 2.38 inwg will then open the damper.

Barometric Pressure Relief Damper William A. Greco [email protected] June 2009

Page 9 of 9 Trident3

Conclusion: A barometric relief damper that is not electrically or electronically controlled must have it’s axle offset from from it’s centerline to work and be properly statically balanced. There are manual barometric relief damper’s sold today that have their axles on the centerline of the damper, their usefullness is in doubt. The mathematics shown in this report show how to build and balance a manual barometric pressure relief damper.

William A. Greco 2404 Greensward N. Warrington, Pa. 18976 References: Machinery’s Handbook 17th Edition 1964 The Engineer’s Manual, Ralph G. Hudson, John wiley and Sons 1917 Marks Standard Handbook For Mechanical Engineers, 7th Edition, McGraw-Hill 1967 Architectural Graphic Standards, Ramsey and Sleeper, 6th Edition, 1970